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Charge a Li-Ion battery from a anemometer

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netrocos

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Hi guys,

I am trying to charge this Li-Ion battery (https://www.sparkfun.com/products/339) with a small dc generator.

It is attached to an anemometer and i can get 100 mV and 20 mA in a normal windy day.

I have this dc/dc converter to boost the 100mV to 4.1V so i can charge the battery (https://www.linear.com/product/LTC3108)

What i need now is to manage the battery so it won't overcharge and don't let it discharge completely.

This circuit is to power an Arduino Fio, so i need low power.

Can someone help me?
 

Your power levels are low enough that you can use a simple zener diode regulator. Select a value equal to the safe voltage limit for your battery.

It may not be easy to obtain the exact value you need. Try stringing together a series of zeners/ diodes/ led's. The TL431 is ideal for this purpose because it works like an adjustable zener.
 

100mV X 20mA = 2mW
Substract at least 20% for the conversion and another 20% for the battery's efficiency. That will leave you with slightly above 1.5mW.

Will that be enough?
The battery is 1000mA/h X 3.7V or 3700mW/h. It will take you in excess of 100 days to charge it fully.
And that is without any load.

EDIT: After reading the converter's datasheet, I noticed that I overestimated its efficiency.....it will take much longer to recharge the battery.
 
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My calculation might be wrong but shows it will take 2100399 hours to fully charge the battery when the wind is blowing continuously full blast and the battery is not powering anything while charging. That is 240 years!
How tiny is your anemometer?

A zener diode will not work at the extremely low current from the voltage boosting IC.

EDIT: Yeah, my calculation was wrong. Charging will take about 4200 hours or almost 6 months continuously.
 
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The missing info is your wind generator power vs air speed and stall threshold. Then compute time to charge up 4Wh. (1Ah*4V~) vs wind speed. Unless RPM is regulated separately from boost DC-DC regulator for optimum power transfer , expect far less than rated power. Better to use higher voltage gen and Buck current booster with MPPT control

Back to the drawing board.
 

Maybe this is similar to the thread about a guy wanting to light the lights on his bicycle with a computer fan as a windmill generator and it was charging a battery. After pedalling into the wind all day then the LED lights would light for about only 20 seconds.
 

I can get 150 mV and 150 mA with 1 impulse... The arduino Fio has very little consumes, so i don't really need to charge the battery completely, i only need to collect some energy through a long period of time to keep the it working in sleep mode and some peak points to get information.

Imagine i would get 150 mV and 150 mA in a constant source. It would give me 22.5 mW of energy. To boost that tension to 4.1 V, what current would i get out of the step up converter?
 

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